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diff --git a/Principles_of_Power_System/chapter14_1.ipynb b/Principles_of_Power_System/chapter14_1.ipynb new file mode 100644 index 00000000..9abb2d75 --- /dev/null +++ b/Principles_of_Power_System/chapter14_1.ipynb @@ -0,0 +1,746 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0910cfded5ae40d0ea2b9576604d8007ec8328215058593fcec218762063deff"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14: A.C. Distribution"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.1, Page Number: 359"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration:\n",
+ "l = 300 #line length(m)\n",
+ "magI1 = 100 #current at load 1(A)\n",
+ "pf1 = 0.707 #power factor at load 1\n",
+ "l1 = 200 #line length till load 1(m)\n",
+ "magI2 = 200 #current drawn at load 2(A)\n",
+ "pf2 = 0.8 #power factor at laod 2\n",
+ "l2 = 300 #line length till load 2(m)\n",
+ "R = 0.2 #total resistance of line(ohm/km)\n",
+ "X = 0.1 #total reactance of the line(ohm/km)\n",
+ "\n",
+ "#Calculation:\n",
+ "phy1 = math.acos(pf1)\n",
+ "phy2 = math.acos(pf2)\n",
+ "Zac = (R+X*1j)*200/1000 #Impedance of section AC(ohm)\n",
+ "Zcb = (R+X*1j)*100/1000 #Impedance of section CB(ohm)\n",
+ "#Taking voltage at the far end B as the reference vector, we have\n",
+ "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n",
+ "I1 = magI1*(pf1-1j*math.sin(phy1)) #Load current at point C(A)\n",
+ "Icb = I2 #A\n",
+ "Iac = I1+I2 #A\n",
+ "Vcb = Icb*Zcb #V\n",
+ "Vac = Iac*Zac #V\n",
+ "V = Vac+Vcb #Voltage drop in the distributor(V)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage drop in the distributor is\",round(abs(V),2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage drop in the distributor is 17.85 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.2, Page Number: 359"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration:\n",
+ "l = 2000 #line length(m)\n",
+ "magI1 = 80 #current at load 1(A)\n",
+ "pf1 = 0.9 #power factor at load 1\n",
+ "l1 = 1000 #line length till load 1(m)\n",
+ "magI2 = 120 #current drawn at load 2(A)\n",
+ "pf2 = 0.8 #power factor at laod 2\n",
+ "l2 = 2000 #line length till load 2(m)\n",
+ "R = 0.05 #total resistance of line(ohm/km)\n",
+ "X = 0.1 #total reactance of the line(ohm/km)\n",
+ "magVb = 230 #voltage maintained at point B(V)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "phy1 = math.acos(pf1)\n",
+ "phy2 = math.acos(pf2)\n",
+ "Zac = (R+X*1j)*1000/1000 #Impedance of section AC(ohm)\n",
+ "Zcb = (R+X*1j)*1000/1000 #Impedance of section CB(ohm)\n",
+ "#Taking voltage at the far end B as the reference vector, we have\n",
+ "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n",
+ "I1 = magI1*(pf1-1j*math.sin(phy1)) #Load current at point C(A)\n",
+ "Icb = I2 #A\n",
+ "Iac = I1+I2 #A\n",
+ "Vcb = Icb*Zcb #V\n",
+ "Vac = Iac*Zac #V\n",
+ "V = Vac+Vcb+magVb*(1+0j) #Voltage drop in the distributor(V)\n",
+ "theta = math.atan(V.imag/V.real)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Voltage drop in the distributor is\",round(abs(V),2),\"V\"\n",
+ "print \"(ii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage drop in the distributor is 261.67 V\n",
+ "(ii)The phase difference between Va and Vb is 3.83 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.3, Page Number: 360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration:\n",
+ "magI1 = 100 #current at load 1(A)\n",
+ "pf1 = 0.6 #power factor at load 1\n",
+ "magI2 = 100 #current drawn at load 2(A)\n",
+ "pf2 = 0.8 #power factor at laod 2\n",
+ "R = 0.1 #total resistance of line(ohm/km)\n",
+ "X = 0.15 #total reactance of the line(ohm/km)\n",
+ "magVb = 200 #voltage maintained at point B(V)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "phy1 = math.acos(pf1)\n",
+ "phy2 = math.acos(pf2)\n",
+ "Zam = (R+X*1j) #Impedance of section AM(ohm)\n",
+ "Zmb = (R+X*1j) #Impedance of section MB(ohm)\n",
+ "#Taking voltage at the far end B as the reference vector, we have\n",
+ "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n",
+ "Imb = I2 \n",
+ "Vb = magVb*(1+0j) #V\n",
+ "Vmb = Imb*Zmb #V\n",
+ "Vm = Vb+Vmb #V\n",
+ "alpha = math.atan(Vm.imag/Vm.real) #V\n",
+ "#The load current I1 has a lagging p.f. of 0\u00b76 w.r.t. VM. It lags \n",
+ "#behind Vm by an angle phy1.\n",
+ "#Phase angle between I1 and Vb\n",
+ "phy11 = phy1-alpha\n",
+ "I1 = magI1*(math.cos(phy11)-math.sin(phy11)*1j) #A\n",
+ "Iam = I1+I2 #A\n",
+ "Vam = Iam*Zam #V\n",
+ "Va = Vm+Vam #V\n",
+ "theta = math.atan(Va.imag/Va.real)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i)Voltage at mid-point is\",round(abs(Vm),1),\"V\"\n",
+ "print \"(ii) Sending end voltage Va is\",round(abs(Va),2),\"V\"\n",
+ "print \"(iii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)Voltage at mid-point is 217.1 V\n",
+ "(ii) Sending end voltage Va is 252.33 V\n",
+ "(iii)The phase difference between Va and Vb is 3.07 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.4, Page Number: 362"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Zab = 1+1j #ohm\n",
+ "Zbc = 1+2j #ohm\n",
+ "Zac = 1+3j #ohm\n",
+ "Ib = 20 #load current at B(A)\n",
+ "pfb = 0.8 #power factor at A\n",
+ "Ic = 15 #load current at C(A)\n",
+ "pfc = 0.6 #power factor at B\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "Iab = Ib*(pfb-1j*math.sin(math.acos(pfb))) #Current in section AB(A)\n",
+ "Iac = Ic*(pfc-1j*math.sin(math.acos(pfc))) #Current in section AB(A)\n",
+ "Vab = Iab*Zab #Voltage drop in section AB(V)\n",
+ "Vac = Iac*Zac #Voltage drop in section AC(V)\n",
+ "#point B is at higher potential than point C. The p.d. between B and C\n",
+ "#is Thevenin\u2019s equivalent circuit e.m.f. Eo i.e.\n",
+ "Eo = Vac-Vab #volt\n",
+ "Zo = Zab+Zac #Thevenin\u2019s equivalent impedance(ohm)\n",
+ "Ibc = Eo/(Zo+Zbc) #A\n",
+ "Iab1 = Iab+Ibc #A\n",
+ "Iac1 = Iac-Ibc #A\n",
+ "Ia = Iab+Iac #Current fed at A(A)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The total current fed at A is\",Ia,\"A\"\n",
+ "print \"\\nCurrent in AB is\",Iab1.real+round(Iab1.imag,2)*1j,\"A\"\n",
+ "print \"\\nCurrent in BC is\",Ibc.real+round(Ibc.imag,2)*1j,\"A\"\n",
+ "print \"\\nCurrent in AC is\",Iac1.real+round(Iac1.imag,2)*1j,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total current fed at A is (25-24j) A\n",
+ "\n",
+ "Current in AB is (18.6-13.53j) A\n",
+ "\n",
+ "Current in BC is (2.6-1.53j) A\n",
+ "\n",
+ "Current in AC is (6.4-10.47j) A\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.5, Page Number: 363"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "l = 1000 #line length(m)\n",
+ "magI1 = 5 #current at load 1(A)\n",
+ "pf1 = 0.8 #power factor at load 1\n",
+ "l1 = 600 #line length till load 1(m)\n",
+ "pf2 = 0.85 #power factor at motor load B\n",
+ "Po = 10 #power output at B(H.P)\n",
+ "n = 0.9 #efficiency\n",
+ "l2 = 400 #line length till load 2(m)\n",
+ "R = 1 #total resistance of line(ohm/km)\n",
+ "X = 0.5 #total reactance of the line(ohm/km)\n",
+ "magVb = 400 #voltage maintained at point B(V)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "Zac = (R+X*1j)*l1/l #ohm\n",
+ "Zcb = (R+X*1j)*l2/l #ohm\n",
+ "magVbp = magVb/3**0.5 #volt per phase\n",
+ "Vbp = magVbp*(1+0j) #V\n",
+ "magIb = Po*746/(3**0.5*magVb*pf2*n) #Line current at B(A)\n",
+ "magI2p = magIb\n",
+ "I2p = magI2p*(pf2-1j*math.sin(math.atan(pf2))) #A\n",
+ "I1p = magI1*(pf1-1j*math.sin(math.atan(pf1))) #A\n",
+ "Iac = I1p+I2p #Current in section AC(A)\n",
+ "Icb = I2p #Current in section CB(A)\n",
+ "Vcb = Icb*Zcb #V\n",
+ "Vac = Iac*Zac #V\n",
+ "Va = Vbp+Vcb+Vac #V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Line voltage at A is\",round(abs(Va)*3**0.5),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Line voltage at A is 434.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.6, Page Number: 364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "magVa = 11000 #volt\n",
+ "magIb = 50 #load current at B(A)\n",
+ "pf2 = 0.8 #power factor(lagging)\n",
+ "magIc = 120 #load current at C(A)\n",
+ "pf3 = 1.0 #power factor(lagging)\n",
+ "magId = 70 #load current at D(A)\n",
+ "pf4 = 0.866 #power factor(lagging)\n",
+ "Zab = 1+0.6j #ohm\n",
+ "Zbc = 1.2+0.9j #ohm\n",
+ "Zcd = 0.8+0.5j #ohm\n",
+ "Zda = 3+2j #ohm\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#Let current in section AB be (x + j*y).\n",
+ "x,y = symbols('x,y')\n",
+ "Iab = x+1j*y\n",
+ "x,y = symbols('x,y')\n",
+ "Ibc = Iab-magIb*(pf2-math.sin(math.atan(pf2))) #A\n",
+ "Icd = (x-40+1j*(y+30))-(120+0j) #A\n",
+ "Ida = ((x-160)+1j*(y+30))-(70*(0.866-0.5j)) #A\n",
+ "Vab = Iab*Zab #Drop in section AB(V)\n",
+ "Vbc = Ibc*Zbc #Drop in section BC(V)\n",
+ "Vcd = Icd*Zcd #Drop in section CD(V)\n",
+ "Vda = Ida*Zda #Drop in section DA(V)\n",
+ "# Vab+Vbc+Vcd+Vda = 0\n",
+ "# As the real and imaginary parts have to be separately zero,\n",
+ "# 6*x-4*y-1009.8 = 0\n",
+ "# 4*x+6*y-302.2 = 0\n",
+ "x1 = solve(6*x-4*y-1009.8,x)[0]\n",
+ "y1 = round(solve(4*x1+6*y-302.2,y)[0],1)\n",
+ "x11 = round(solve(6*x-4*y1-1009.8,x)[0],1)\n",
+ "#now putting the values of x11 and y1 in above equationa,\n",
+ "Iab1 = x11+1j*y1 #A\n",
+ "Ibc1 = (x11-40)+1j*(y1+30) #A\n",
+ "Icd1 = (x11-40+1j*(y1+30))-(120+0j) #A\n",
+ "Ida1 = ((x11-160)+1j*(y1+30))-(70*(0.866-0.5j)) #A\n",
+ "magVap = round(magVa/3**0.5) #Voltage at supply end A(V)\n",
+ "Vb = magVap*(1+0j)-Iab1*Zab #Voltage at station B(V/phase)\n",
+ "Vc = Vb-Ibc1*Zbc #Voltage at station C(V/phase)\n",
+ "Vd = Vc-Icd1*Zcd #Voltage at station D(V/phase)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Current in section AB is\",Iab1,\"A\"\n",
+ "print \"Current in section BC is\",Ibc1,\"A\"\n",
+ "print \"Current in section CD is\",Icd1,\"A\"\n",
+ "print \"Current in section DA is\",Ida1,\"A\"\n",
+ "print \"Voltage at A is\",magVap*(1+0j),\"V/phase\"\n",
+ "print \"Voltage at B is\",Vb,\"V/phase\"\n",
+ "print \"Voltage at C is\",Vc,\"V/phase\"\n",
+ "print \"Voltage at D is\",Vd,\"V/phase\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in section AB is (139.8-42.8j) A\n",
+ "Current in section BC is (99.8-12.8j) A\n",
+ "Current in section CD is (-20.2-12.8j) A\n",
+ "Current in section DA is (-80.82+22.2j) A\n",
+ "Voltage at A is (6351+0j) V/phase\n",
+ "Voltage at B is (6185.52-41.08j) V/phase\n",
+ "Voltage at C is (6054.24-115.54j) V/phase\n",
+ "Voltage at D is (6064-95.2j) V/phase\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.7, Page Number: 368"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Pr = 10 #load connected to line R(kW)\n",
+ "Py = 8 #load connected to line Y(kW)\n",
+ "Pb = 5 #load connected to line B(kW)\n",
+ "Vl = 400 #line voltage(V)\n",
+ "#the loads are non-reactive.\n",
+ "\n",
+ "#Calculation:\n",
+ "Vp = round(Vl/3**0.5) #phase voltage(V)\n",
+ "Ir = Pr*1000/Vp #A\n",
+ "Iy = Py*1000/Vp #A\n",
+ "Ib = Pb*1000/Vp #A\n",
+ "\n",
+ "#Resolving the three currents along x-axis and y-axis, we have,\n",
+ "Ih = Iy*math.cos(math.pi/6)-Ib*math.cos(math.pi/6) #Resultant horizontal component(A)\n",
+ "Iv = Ir-Iy*math.cos(math.pi/3)-Ib*math.cos(math.pi/3) #Resultant vertical component(A)\n",
+ "In = (Ih**2+Iv**2)**0.5 #current in neutral wire(A)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Ir =\",round(Ir,1),\"A\"\n",
+ "print \" Iy =\",round(Iy,1),\"A\"\n",
+ "print \" Ib =\",round(Ib,2),\"A\"\n",
+ "print \"(ii) Current in neutral wire is\",round(In,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Ir = 43.3 A\n",
+ " Iy = 34.6 A\n",
+ " Ib = 21.65 A\n",
+ "(ii) Current in neutral wire is 18.9 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.8, Page Number: 369"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Vl = 400 #line voltage(V)\n",
+ "Vp = 230 #voltage across lamp(V)\n",
+ "I1 = 70 #current in load RN(A)\n",
+ "I2 = 84 #current in load YN(A)\n",
+ "I3 = 33 #current in load BN(A)\n",
+ "Im = 200 #current taken by the motor(A)\n",
+ "pf = 0.2 #power factor(lagging)\n",
+ "\n",
+ "#Calculation:\n",
+ "#Lamp load alone:\n",
+ "Ih = I2*math.cos(math.pi/6)-I3*math.cos(math.pi/6) #Resultant H-component(A)\n",
+ "Iv = I1-I3*math.cos(math.pi/3)-84*math.cos(math.pi/3) #Resultant V-component(A)\n",
+ "In = math.sqrt(Ih**2+Iv**2) #Neutral current(A)\n",
+ "\n",
+ "\n",
+ "#Both lamp load and motor load:\n",
+ "Ir = Im*pf #Active component of motor current(A)\n",
+ "Ix = Im*math.sin(math.acos(pf)) #Reactive component of motor current(A)\n",
+ "IR = ((Ir+I1)**2+Ix**2)**0.5 #A\n",
+ "IY = ((Ir+I2)**2+Ix**2)**0.5 #A\n",
+ "IB = ((Ir+I3)**2+Ix**2)**0.5 #A\n",
+ "P = Vp*(I1+I2+I3)*1 #Watt #( cos phy_L = 1)\n",
+ "Pm = 3**0.5*Vl*Im*pf #Power supplied to motor(W)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Lamp load alone: neutral curent is\",round(In,2),\"A\"\n",
+ "print \"\\nWhen Both lamp load and motor load is present:\"\n",
+ "print \"The current components are:\"\n",
+ "print \"Neutral current is\",round(In,2),\"A\"\n",
+ "print \"IR =\",round(IR,1),\"A;\\tIY =\",round(IY,0),\"A;\\tIB =\",round(IB,2),\"A\"\n",
+ "print \"Power supplied to the lamp is\",P,\"W\"\n",
+ "print \"Power supplied to the motor is\",round(Pm),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lamp load alone: neutral curent is 45.64 A\n",
+ "\n",
+ "When Both lamp load and motor load is present:\n",
+ "The current components are:\n",
+ "Neutral current is 45.64 A\n",
+ "IR = 224.7 A;\tIY = 232.0 A;\tIB = 209.11 A\n",
+ "Power supplied to the lamp is 43010 W\n",
+ "Power supplied to the motor is 27713.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.9, Page Number: 370"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Prn = 20 #kW\n",
+ "pf1 = 1 #power factor of loaf RN\n",
+ "kVAyn = 28.75 #kVA of load YN\n",
+ "kVAbn = 28.75 #kVA of load BN\n",
+ "pf2 = 0.866 #power factor of laod YN & BN each.(lagging)\n",
+ "Vl = 400 #line voltage(V)\n",
+ "Vp = 230 #phase voltage(V)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "phy1 = math.acos(pf1)\n",
+ "phy2 = math.acos(pf2)\n",
+ "phy3 = phy2\n",
+ "Ir = Prn*1000/Vp #A\n",
+ "Iy = kVAyn*1000/Vp #A\n",
+ "Ib = kVAbn*1000/Vp #A\n",
+ "Ih = Ir-Iy*math.cos(phy2)-Ib*math.cos(phy2) #A\n",
+ "Iv = 0+Iy*math.sin(phy2)-Iy*math.sin(phy3) #A\n",
+ "In = math.sqrt(Ih**2+Iv**2) #A\n",
+ "#When load from B to N removed.:\n",
+ "#When the load from B to N is removed, the various line currents are:\n",
+ "#Ir in phase with Vrn; Iy lagging by 30 deg.; Ib = 0.\n",
+ "Ir1 = Ir\n",
+ "Iy1 = Iy; Ib1 = 0 #A\n",
+ "Ih1 = Ir1-Iy1*math.cos(math.pi/6) #A\n",
+ "Iv1 = 0-Iy1*math.sin(math.pi/6) #A\n",
+ "In1 = math.sqrt(Ih1**2+Iv1**2) #A\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"When no changes were made, the various currents are:\"\n",
+ "print \"Ir =\",round(Ir,2),\"A;\\tIy =\",Iy,\"A;\\tIb =\",Ib,\"A;\\tIn =\",round(In,2),\"A\"\n",
+ "print \"\\nWhen load from B to N removed, the various currents are:\"\n",
+ "print \"Ir =\",round(Ir1,2),\"A;\\tIy =\",Iy1,\"A;\\tIb =\",Ib1,\"A;\\tIn =\",round(In1,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When no changes were made, the various currents are:\n",
+ "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 125.0 A;\tIn = 129.54 A\n",
+ "\n",
+ "When load from B to N removed, the various currents are:\n",
+ "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 0 A;\tIn = 66.03 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.10, Page Number: 371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import cmath\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Vl = 400 #line voltage(V)\n",
+ "Vp = 230 #phase voltage(V)\n",
+ "Ir = 30 #load current at R-phase(A)\n",
+ "pf1 = 0.866 #power factor for R-phase(lagging)\n",
+ "Iy = 30 ##load current at Y-phase(A)\n",
+ "pf2 = 0.866 #power factor for R-phase(lagging)\n",
+ "Ib = 30 ##load current at R-phase(A)\n",
+ "pf3 = 1.0 #power factor for R-phase(lagging)\n",
+ "R = 0.2 #resistance of each line conductor(ohm)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "phy1 = math.acos(pf1)\n",
+ "phy2 = math.acos(pf2)\n",
+ "phy3 = math.acos(pf3)\n",
+ "\n",
+ "VR = Vp*(1+0j) #V\n",
+ "VY = Vp*(math.cos(-2*math.pi/3)+math.sin(-2*math.pi/3)) #V\n",
+ "VB = Vp*(math.cos(2*math.pi/3)+math.sin(-2*math.pi/3)) #V\n",
+ "\n",
+ "#the line currents can be expressed as :\n",
+ "IR = cmath.rect(30,-math.pi/6) #A\n",
+ "IY = cmath.rect(30,-math.pi/2) #A\n",
+ "IB = cmath.rect(30,2*math.pi/3) #A\n",
+ "IN = IR+IY+IB #A\n",
+ "\n",
+ "#Since, the area of X-section of neutral is half of any line conductor.\n",
+ "Rn = 2*R #resistance of neutral(ohm)\n",
+ "#ER = VR + Drop in R phase + Drop in neutral\n",
+ "ER = VR+R*IR+IN*2*R #V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The supply end voltage for R phase is\",round(ER.real,3)+1j*round(ER.imag,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The supply end voltage for R phase is (239.588-10.608j) V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.11, Page Number: 371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Vl = 400 #line voltage(V)\n",
+ "Vp = 230 #phase voltage(A)\n",
+ "Pln = 100 #load connected b/n LN(W)\n",
+ "Pyn = 150 #load connected b/n YN(W)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#before disconnecting the neutral wire,\n",
+ "R1 = Vp**2/Pln #Resistance of lamp L1(ohm)\n",
+ "R2 = Vp**2/Pyn #Resistance of lamp L2(ohm)\n",
+ "\n",
+ "#When the neutral wire is disconnected,\n",
+ "EL = 400 #V\n",
+ "I = EL/(R1+R2) #A\n",
+ "V1 = I*R1 #Voltage across lamp L1(V)\n",
+ "V2 = I*R2 #Voltage across lamp L2(V)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The voltage across the lamps are:\"\n",
+ "print \"Lamp 1, Voltage =\",V1,\"V ;\\tLamp 2, voltage =\",V2,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The voltage across the lamps are:\n",
+ "Lamp 1, Voltage = 240.0 V ;\tLamp 2, voltage = 160.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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